Question

Solve for n:

(4^n)^3 (2^n + 2) = (4^2)^n – 1

Answer 1

(n-1)!/(n-2)!+(n-2)!/(2!(n-4)!)=4
(((n-2)!)^2 + 2(n-4)! (n-1)!) / (2(n-4)! (n-2)!) = 4
1/2*(n^2 - 3n + 4) = 4
n^2 - 3n - 4 = 0
(n + 1)(n - 4) = 0
n = -1 => reject
n = 4